Reaction of acetamide with bromine and KOH gives

Options

A. `CH_(3)COOH`
B. `CH_(3)CH_(2)NH_(2)`
C. `CH_(3)COONH_(4)`
D. `CH_(3)NH_(2)`.

Explanation

To solve the question regarding the reaction of acetamide with bromine (Br2) and potassium hydroxide (KOH), we can follow these steps:

### Step 1: Identify the Reactants
Acetamide is an amide with the chemical formula CH3CONH2. The other reactants are bromine (Br2) and potassium hydroxide (KOH).

### Step 2: Understand the Reaction Conditions
The reaction involves bromine in the presence of a strong base (KOH). This indicates that we may be dealing with a halogenation reaction, specifically a haloform reaction.

### Step 3: Write the Reaction
When acetamide reacts with bromine and KOH, the bromine acts as a halogenating agent. The reaction can be summarized as follows:
\[ \text{CH}_3\text{CONH}_2 + \text{Br}_2 + \text{KOH} \rightarrow \text{CH}_3\text{COOK} + \text{NH}_2\text{Br} + \text{H}_2\text{O} \]

### Step 4: Identify the Products
The products of the reaction are:
1. Potassium acetate (CH3COOK)
2. Bromoamine (NH2Br)
3. Water (H2O)

### Step 5: Conclusion
Thus, the reaction of acetamide with bromine and KOH yields potassium acetate, bromoamine, and water.

### Final Answer:
The reaction of acetamide with bromine and KOH gives potassium acetate (CH3COOK), bromoamine (NH2Br), and water (H2O).

---